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Normal force changes

  1. Apr 1, 2010 #1
    Hi guys, I am confused about how the normal force exerted on an object changes depending on the situation.

    Let's say an object weighs 10 N at rest. The normal force here is 10 N as well since that is by how much the object is pushing down on the surface.

    When someone tries to pull it upward with a 6 N force, it "relieves" some of the normal force. Since the object is now only pushing down on the surface with 4 N, the normal force is 4 N.

    So the proper set-up for an equation here is F(normal) = F(gravity) - F(upward)

    But in an elevator, let's say the elevator is accelerating upward at 2 m/s^2. If a person were inside the elevator, why doesn't this "relieve" some of their normal force as well? Instead, I have been told that the proper equation for this situation is F(upward) = F(normal) - F(gravity). Rearranging that equation, F(normal) = F(upward) + F(gravity). So now the normal force is actually much greater as a result of an upward force.

    Why is this so?
     
  2. jcsd
  3. Apr 1, 2010 #2
    action- reaction , if the elevator is accelerating upward this adds to the force , like you would feel heavier going up. And in relativity there is no difference between being accelerated by a rocket or being in a gravitational field .
     
  4. Apr 1, 2010 #3

    tiny-tim

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    Welcome to PF!

    Hi Tokspor! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Because you're using "F(upward)" to mean two different things

    in the first case, it's a separate (third) applied force, but in the second case it's the total of the (two) forces. :wink:

    Let's apply good ol' Newton's second law …

    in the first case, a = 0, and so all the forces must add to 0 …

    F(normal) + F(gravity) + F(applied) = 0,

    ie F(normal) = mg - F(applied),​

    ('cos gravity is downward :wink:)

    but in the second case, a = 2,

    F(normal) + F(gravity) = 2,

    ie F(normal) = mg + 2,​

    … see? no applied force!! :biggrin:
     
  5. Apr 1, 2010 #4

    Doc Al

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    The proper set-up in both cases is to use Newton's 2nd law:

    ΣF = ma

    F(normal) - F(gravity) = ma

    When the acceleration is zero, the normal force equals your weight. If the elevator is accelerating upward (thus a > 0), the normal force is greater than your weight--you feel heavier. (The normal force must not only support your weight but accelerate you.)

    Edit: While I was goofing off, tiny-tim beat me to it! :cool:
     
  6. Apr 1, 2010 #5
    This is because in this elevator scenario, the normal force is the source of this "upward force" too. If suppose a huge guy infront holds you by your neck and lifts you, the "upward force" is provided by *his* hands. In that case, the normal force gets some help and is relieved as you hoped. (Also hope he releases you soon.)
     
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